1.3 Dot Product - Kennesaw State University

1.3. DOT PRODUCT

19

1.3 Dot Product

1.3.1 De...nitions and Properties

The dot product is the ...rst way to multiply two vectors. The de...nition we will give below may appear arbitrary. But it is not. It is motivated by applications, in particular projections.

De...nition 34 (Dot Product) The dot product, also called inner product,

is denoted with the symbol : We can only take the dot product of two vectors

having the same dimension. The result is a number, obtained by multiplying the

corresponding coordinates of the two vectors, then adding all the products. If

~u = ha; bi and ~v = hc; di are two vectors, then ~u ~v = ac + bd. You will note that

the answer is a number, not a vector. We have a similar de...nition for vectors in

higher dimensions. In general, if ~u = hx1; x2; Pn

then ~u ~v = xiyi.

i=1

; xni and ~v = hy1; y2;

; yni,

Example 35 If ~u = h1; 1; 1i and ~v = h1; 2; 3i then ~u ~v = (1) (1) + (1) (2) + (1) (3) = 6.

Proposition 36 The dot product satis...es the following properties: 1. ~u ~u = k~uk2 2. ~u ~0 = ~0 ~u = 0 3. ~u ~v = ~v ~u 4. ~u ~v = (~u ~v) for any scalars and . 5. ~u (~v + w~ ) = ~u ~v + ~u w~ 6. If is the angle (between 0 and ) between two non-zero vectors ~u and ~v, then ~u ~v = k~uk k~vk cos . 7. ~u ? ~v if and only if ~u ~v = 0. Again, we assume that ~u and ~v are two non-zero vectors. 8. j~u ~vj k~uk k~vk

Proof. We give an idea of a proof for some of these properties.

1. We see why this is true for a 3-D vector. A general proof is similar. Suppose that ~u = ha; b; ci. Then, by de...nition

~u ~u = a2 + b2 + c2

Also by de...nition,

p k~uk = a2 + b2 + c2

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CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE

Thus

k~uk2 = a2 + b2 + c2 = ~u ~u

2. Left as an exercise.

3. Left as an exercise.

4. Left as an exercise.

5. Left as an exercise.

6. Consider ...gure 6. Using the law of cosines, we see that k~u ~vk2 = k~uk2 + k~vk2 2 k~uk k~vk cos

Thus 2 k~uk k~vk cos

2 k~uk k~vk cos

= k~u ~vk2 + k~uk2 + k~vk2 = (~u ~v) (~u ~v) + ~u ~u + !v !v = (~u ~u 2~u ~v + !v !v ) + ~u ~u + !v !v = ~u ~u + 2~u ~v !v !v + ~u ~u + !v !v

= 2~u ~v

Hence

~u ~v = k~uk k~vk cos

7. If the vectors are perpendicular, then the angle between them is 90 . Thus, cos = 0. The result follows from property 6. Conversely, if ~u and ~v are two non-zero vectors such that ~u ~v = 0 then k~uk k~vk cos = 0. Since k~uk 6= 0 and k~vk 6= 0 (~u and ~v are two non-zero vectors), it must be that cos = 0 thus = 90 hence ~u ? ~v.

1.3. DOT PRODUCT

21

8. From 6, we see that since jcos j 1:

j~u ~vj = jk~uk k~vk cos j = k~uk k~vk jcos j k~uk k~vk

Remark 37 We can also use the dot product to prove the triangle inequality k~u + ~vk k~uk + k~vk. Using the fact that ~u ~u = k~uk2, we have

k~u + ~vk2 = (~u + ~v) (~u + ~v)

= ~u ~u + 2~u ~v + ~v ~v using properties of the dot product

= k~uk2 + 2~u ~v + k~vk2

k~uk2 + 2 j~u ~vj + k~vk2 property of absolute value

k~uk2 + 2 k~uk k~vk + k~vk2 by property 8 above

k~u + ~vk2

(k~uk + k~vk)2

Since both k~u + ~vk and k~uk + k~vk are positive, we can take the square root on both side of the last equality to obtain the result.

Remark 38 Property 6 in proposition 36 can also be used to determine how close two non-zero vectors are to pointing in the same direction. Given two non-zero vectors ~u and ~v, consider P, the plane passing through the origin and perpendicular to ~u. ~v will be on the same side of P as ~u if and only if ~u ~v > 0 because in that case, the angle between the two vectors will be less than 90 and its cosine will be positive.

Remark 39 Property 7 in proposition 36 is also very important, it gives us a

quick way to check if two vectors are perpendicular. Another term for perpendicular is orthogonal. We de...ne the zero vector ~0 = (0; 0; :::) to be orthogonal

to every vector.

Remark 40 It is important to understand that when we talk about the angle

between two vectors, we are talking about the smallest positive angle between

them in other words an angle such that 0

.

We illustrate some of these properties with examples.

Example 41 Suppose that the length of a vector !u is 2, the length of a vector !v is 3 and they make an angle of 135 as shown in ...gure 1.11. Find ~u !v .

~u !v = k~uk k~vk cosp135!

= (2) (3)

2

2

p = 32

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CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE

Figure 1.11: Compute !u !v if k!u k = 2 and k!v k = 3

Example 42 What is the angle in degrees between !u = h1; 1; 1i and !v = h2; 1; 0i? From the formula ~u ~v = k~uk k~vk cos , we can compute cos and therefore . First, let us compute the various quantities needed.

~u ~v = (2) (1) + (1) (1) + (1) (0) =3

p k~uk = 12 + 12 + 12

p =3

k!v k

=

p 22 + 12

p

=5

It follows that

= cos 1 ~u ~v k~uk k~vk

= cos 1 p3 15

= 39:23

1.3. DOT PRODUCT

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Figure 1.12: proj ~a~b and orth~a~b

1.3.2 Projections

It often happens that we need to decompose a vector into components that are parallel and perpendicular to a given vector. Or, as shown in ...gure 1.12, given two non-zero vectors ~a and ~b, one needs to ...nd the projection of ~b onto ~a and the projection of ~b onto a vector perpendicular to ~a. Let us denote proj ~a~b the projection of ~b onto ~a, comp~a~b the component of ~b along ~a (also known as the scalar projection of ~b onto ~a) and orth~a~b the projection of ~b onto a vector perpendicular to ~a.Let us call the angle between ~a and ~b. Then, we have:

cos = comp~a~b ~b

Therefore

comp~a~b = ~b cos

Using Property 6 in proposition 36, we can write

~b ~a ~b comp~a~b = k~ak ~b

(1.4)

It follows that

comp~a~b

=

~a ~b k~ak

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CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE

To ...nd proj ~a~b, we simply multiply comp~a~b by a unit vector in the direction of ~a

~a. Such a vector is and therefore k~ak

proj ~a~b

=

~a ~b k~ak2 ~a

(1.5)

Finally, orth~a~b = ~b proj ~a~b and therefore

orth~a~b = ~b

~a ~b k~ak2 ~a

(1.6)

Example 43 Find the scalar projection and vector projection of !b = h1; 1; 2i onto !a = h 2; 3; 1i. Let us ...rst compute the various quantities involved.

!a

! b

=

2+3+2

=3

q j!a j = ( 2)2 + 32 + 12

p = 14

Therefore and

comp~a~b

=

p3 14

proj ~a~b

=

3 14

h

2; 3; 1i

1.3.3 Applications

The dot product has many applications including:

1. Use it to see if two non-zero vectors are perpendicular. Recall that two non-zero vectors are perpendicular if and only if their dot product is 0.

2. Use it to compute the angle between two vectors. From the formula ~u ~v = k~uk k~vk cos , if we are given ~u and ~v, we have

= cos 1 ~u ~v k~uk k~vk

3. Use it to quickly determine if the angle between two vectors is acute (between 0 and 90 degrees) or obtuse (between 90 and 180 degrees). Given ~u and ~v, if the angle between them is acute, then ~u ~v > 0 and if it is obtuse, ~u ~v < 0.

1.3. DOT PRODUCT

25

4. Given a vector and a plane perpendicular to it, use it to determine if a second vector is on the same side of the plane or the opposite side. If the two vectors are on the same side of the plane, the angle between them will be acute. If they are on opposite sides, it will be obtuse. So, we can use the technique describe above.

5. Use it to ...nd the projection of a vector onto another one.

6.

Use it to compute the work done by a force acting on displacing it. In physics, the work W done by a force

a!Fn

object and moving an

othbejeccotmaploonngenttheofd!Fispalalocnemg e!rntbvyectthoer

!r is de...ned magnitude of

to the

be the product of displacement (see

...gure 6). This gives us:

W=

! comp!r F

j!r j

=

!F !r j!r j

j!r j

= !F !r

We illustrate that last application with an example.

Example 44 A crate is displaced by 8m up a ramp by a constant force !F of 200N applied at an angle of 25 degrees to the ramp. Find W , the work done.

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CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE

Let !D be the displacement vector. Then !D = 8 and F = 200:

W = !F !D = !D F cos 25

= (8) (200) cos 25 1450N:m 1450J

Note that in the metric system, distances are in meters (m), forces are in Newton (N), work is expressed in Joules (J).

1.3.4 Problems

1. Prove that the normal to the line ax + by + c = 0 (in 2D) is ha; bi (hint: Find 2 points on the line, deduce what a vector parallel to the line is. From there, ...nd a vector perpendicular to it).

2. For each problem below, ...nd the following:

!v !u , k!v k, k!u k

The cosine of the angle between !u and !v .

comp!v !u proj!v !u

(a) !v = 2!i 4!j + p5!k and !u = 2!i + 4!j (b) !v = 10!i + 11!j 2!k and !u = 3!j + 4!k

(c) !v = 5!j

! 3k

and

!u

=

!i

+

!j

+

! k

(d) !v = 5!i + !j and !u = 2!i + p17!j

p5!k

3. Find the angle between !u = 2!i + !j and !v = !i + 2!j !k . 4. Find the angle between !u = p3!i 7!j and !v = p3!i + !j 2!k .

5. Find the measures of the angles of the triangle whose vertices are A = ( 1; 0), B = (2; 1) and C = (1; 2).

6. Let , , and

be

the

direction

angles

of

!v

=

a!i

+

b!j

+

! ck

(a) Show that cos

=

a

k!v k

,

cos

=

b

k!v k

,

and

cos

=

c

k!v k

and

cos2

+

cos2 + cos2 = 1. These are called the direction cosines.

(b) Show if !v = a!i + b!j + c!k is a unit vector then a, b, c are the

direction cosines.

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